possibility/impossibility of pure Geometric constructions. (1 Viewer)

[SeDaTeD]

Yeah, that's the bit I was thinking you'd be picky on.

 

[who_loves_maths]

Originally Posted by Templar
Read previous threads please. Then you'll realise you can't draw unit chords.

Originally Posted by SeDaTeD
Hmm, what about constructing 3 tangents to the circle, creating a triangle with sides of lenght a, b and c. Then the area of the triangle can be found by Heron's formula then equated with A=rs, where r is the radius of the circle and s = (a+b+c)/2. Then r can be found. Then you can do a bit of island hopping around the circumference with chords of length r, then joining up the opposite ones, which should interset at the centre. But then this method won't please Templar either so I'll leave it at that.

Originally Posted by Templar
Can you even draw a tangent? I'm not sure how you can do so. But good thinking.

apart from the fact that you can't draw a tangent to a circle without being able to construct a right angle (which you can't using just a ruler), can you explain what you mean by "island hopping around the circumference with chords of length 'r' " plz SeDaTeD? do you mean drawing chords of length 'r' in the circle?

if that's what you mean, then wouldn't you not be able to do that due to Templar's earlier comment: "Then you'll realise you can't draw unit chords." ? the fact that you can't inscribe chords of fixed/known length into the circle ?

if i'm missing the point here, then plz tell me.

 

[casebash]

It is not at all clear from previous posts that you cannot draw chords of fixed length inside a circle. You said you can't do trial and error, and this is not. You fix one mark on a circle and rotate until another mark is on the circumference. I admit i didn't read that the ruler had infinite accuracy, i though it had finite. Regardless, with this interpretaion, it is still easy. Draw a random triangle ABC. Figure out using trighow far along BC the angle bisector of A will be. For AB figure out how far along the angle bisector of C will be. Draw these bisectors. They meet at center of circle.

 

[SeDaTeD]

Yes i meant putting chords of length r around the circle. And yes it was also fudgy, but still a contribution. But i do suspect it's probably impossible (possibly because I can't see a way atm).

 

[Templar]

Figure out using trig how far along BC the angle bisector of A will be.

throught pure constructions

No........

 

[who_loves_maths]

Originally Posted by casebash
It is not at all clear from previous posts that you cannot draw chords of fixed length inside a circle. You said you can't do trial and error, and this is not. You fix one mark on a circle and rotate until another mark is on the circumference. I admit i didn't read that the ruler had infinite accuracy, i though it had finite.

hi casebash,

it was Templar who said that you cannot draw lengths of fixed lengths or unit length inside the circle. however, I agree with you that you indeed can do this with a "real-life" ruler - in which case your originally is quite plausible. [which is also why in my last post to SeDaTeD, i quoted Templar rather than taking what he said as a truth.]

But in any case, i suspect what Templar was trying to say more accurately is that in this particular problem we are NOT using such a "real-life" ruler. if you read my explanation at the beginning of this thread you'd know that the definition, here, of the ruler is a STRAIGHTEDGE but with calibrations - this simply means that the problem is essentially the same as that of determining the centre of a circle with a straightedge, but with the exception that you can use the ruler to merely MEASURE lengths or ratios on lines... ie. it does NOT mean you can use it, in the way you did, to artificially draw a line of a fixed length of your will.
in other words, the use of the ruler is passive here, not active.

think of it this way - you can ONLY use the calibrations on the ruler/straightedge to measure out lengths AFTER you have already drawn a (perhaps random) line/chord inside the circle (or outside) that is to be measured or analysed. so you can't use the ruler to measure out a fixed length beforehand and then use that length to trace out a line, like you did.
so in this case Templar is right in saying that your method is similar to the one discussed in this thread previously - in that in order to draw your unit lengthed chord, you will have had to draw a random line through the circle first, and hope that the interval of this line lying inside the circle (the chord) is indeed of one unit length - in which case this is a method based on luck and chance, a "hit-n-miss" solution - and that, like the first solution of SeDaTeD's, is NOT allowed here, as you will have read by now.

so i hope you understand what Templar was trying to say now when he said "Read previous threads please. Then you'll realise you can't draw unit chords."

in anycase, i hope i clarified some things for you here

 

[who_loves_maths]

Originally Posted by casebash
Regardless, with this interpretaion, it is still easy. Draw a random triangle ABC. Figure out using trighow far along BC the angle bisector of A will be. For AB figure out how far along the angle bisector of C will be. Draw these bisectors. They meet at center of circle.

any two angle bisectors of a random triangle do not necessarily meet/intersect at the centre of the its circumcircle at all... it only happens with equilateral triangles.

so by drawing a "random" triangle ABC and using that method, then you are, once again, relying on chance/luck that the random triangle you've constructed is indeed equilateral. this process of construction is unstructured, and therefore invalid as a solution.

additionally, if you go about constructing an equilateral triangle inside the fixed circle using only the ruler, then that would also require you measuring out premeditated lengths (or even angles, which is impossible to do) before you draw the lines in, same as that of your first method, which is, again, not allowed (otherwise, SeDaTeD's second method would have sufficed anyway).

 

[casebash]

Sorry, i got my incircle confused with my circumcircle. I'm getting a bit rusty, i haven't done any serious maths (HSC maths doesn't count as serious) for almost a term. But this doesn't matter, the basic idea of the method (drawing lines on the the sides that will pass through the point).

Draw in a random triangle ABC. You can find its side lengths, so you know know everything about it, including the location of the circumcenter, O(by coordinate bashing). So coordinate bash to get the how far along BC its intersection with AO, call D. Join AD. Do the same with B and AC. Then you have two lines that intersect at the center. But it is an awful method. I'm not going to write the equations because they will get quite long.

 

[who_loves_maths]

^ i'm afraid i'm going to have to agree with you that yes it is an 'awful' method...
you'll need to be MUCH more specific than that if you want that method to be passed off as a legitimate "proof"...
everything might sound 'ok' in a GENERAL outline, but it doesn't mean when you actually get down to doing it that everything will indeed comply - i'm sure you've met many such situations from experience in mathematics. in fact, many serious and challenging maths problems in history have pocured such general drafts of solutions but that which has survived and resisted any serious attempts at solution (eg. Fermat's Last Theorem).

of course, this small problem here is incomparable to the magtitude of FLT; but nonetheless, being a mathematical problem, all the formalisms of a 'proof' must still be strictly followed. eg. you never know, without actually doing out, if there's something in your solution that might, say, require a measurement of any angle or other geometric construction that might be impossible with just a ruler, etc...

anyone knows that the circumcentre of any triangle is the centre of the circle that passes through the vertices of that triangle... so basically anyone can just say "yes, all you have to do is simpy draw a few lines that intersect at the circumcentre, and then you'll have found the centre of the circle. the rest, such as equations, are too compicated and are just "details"..."
- so i'm afraid saying merely that is not satisfactory in this case.


P.S. it is preferrable if, instead of using equations and trig. etc, you could do this problem using, as i said before, pure geometric constructions.

 

[casebash]

Ok, I admit i'm lazy, i try to get people to fill it in for themselves so i don't have to type a lot. Ok, given any two lines you can find the intersection and any two point the line. Given a point and i gradient you know the equation of the line. Finally the gradient of a line perpencicular to a line with a known equation can be calculated. First find area using that semiperimetre formula. Then get the angles with the sin area law. You can then get coordinates of the three points (relative), just construct perpenduclars to the base and componatalise. With coordinates of the vertices, get equations for lines. Then get gradient of lines perpendicular to them. OK, can get coordinates of midpoints. With coordinates of midpoints and gradients perpenduclar, get equations of perpendiclar bisectors. Get intersection at circumcenter. Now Get equation of lines between vertacices and the circumcenter. Get intersections with the sides. Now, you can finally get the points of these intersections, and join them to meet at the circumcentre. Completely obvious, and takes way way way too much effort to type out. Ok, maybe i shouldn't complain too much, but as i said i am lazy. As for pure geometric constructions, well this is a starting point.

 

[casebash]

Ah, got it, a real geometric proof. Find the midpoints of the triangle. Draw a line through two of them to get a line parrallel to the base. Join both the bisection of this and the bisection of the base to get a line through the center. Bisect that.

 

[who_loves_maths]

Originally Posted by casebash
Ok, I admit i'm lazy, i try to get people to fill it in for themselves so i don't have to type a lot. Ok, given any two lines you can find the intersection and any two point the line. Given a point and i gradient you know the equation of the line. Finally the gradient of a line perpencicular to a line with a known equation can be calculated. First find area using that semiperimetre formula. Then get the angles with the sin area law. You can then get coordinates of the three points (relative), just construct perpenduclars to the base and componatalise. With coordinates of the vertices, get equations for lines. Then get gradient of lines perpendicular to them. OK, can get coordinates of midpoints. With coordinates of midpoints and gradients perpenduclar, get equations of perpendiclar bisectors. Get intersection at circumcenter. Now Get equation of lines between vertacices and the circumcenter. Get intersections with the sides. Now, you can finally get the points of these intersections, and join them to meet at the circumcentre. Completely obvious, and takes way way way too much effort to type out. Ok, maybe i shouldn't complain too much, but as i said i am lazy. As for pure geometric constructions, well this is a starting point.

okay, i'm really starting to get annoyed now at the fact that you OBVIOUSLY have NOT read any of the posts in this thread previous to your first...
so i'll go through it with you one last time, plz read carefully this time.

1) the question is: how do you find the centre of a given, fixed, circle using ONLY a ruler and nothing else? where a ruler is a straightedge with calibrations.

2) do the question using geometric constructions, as if you would will a real straightedge, with calibrations, in real life.


Now, here's what's wrong with your above method (esp. the lines in bold quoted above):

1) we don't use the term "gradient" when approaching this problem from a purely geometric point of view since the term "gradient" {esp. in the sense that you used it} implies the use of coordinate geometry - which is *NOT* an allowable approach to this question because you using coordinate geometry does *NOT* let you side-step some of the problems concerning the use of ONLY a ruler...

2) we don't use "equations" of lines when approaching this problem '' '' ... '' '' ... because it implies the use of coordinate geometry - which is *NOT* an allowable approach to this question because you using coordinate geometry does *NOT* let you side-step some of the problems concerning the use of ONLY a ruler...

3) we don't use "coordinates" of lines when approaching this problem because (i) you ONLY have a ruler and nothing else; so knowing the coordinates of a point within the circle will *NOT* help you in anyway to located that point using just a ruler. The circle does *NOT* lie on a grid of infinite accuracy! and (ii) because "coordinates" implies the use of coordinate geometry - which is *NOT* an allowable approach to this question because you using coordinate geometry does *NOT* let you side-step some of the problems concerning the use of ONLY a ruler...

4)

"You can then get coordinates of the three points (relative), just construct perpenduclars to the base and componatalise... " - NO, you cannot construct perpendiculars to the base, because you have nothing to measure the (right) angle with! you ONLY have a ruler!

5)

" ...get equations of perpendiclar bisectors. Get intersection at circumcenter... " - NO, you cannot construct bisectors using ONLY a ruler for you have nothing to measure the angle with.

6) your approach is certainly *NOT* "completely obvious" {as you put it} - because (i) it is illegitimate, and (ii) you have *NOT* explained yourself very well or clearly at all in that post.



so once again, i beg you to either (i) re-read, not re-skim, everything that's been posted and said before your first post carefully, and/or, (ii) *DO NOT* post something that you have not given a lot of thought too {which would also probably be flawed} again until you have FULLY understood the basic requirements that a 'proof' needs to satisfy here for this particular problem.

so until or unless you are very confident of your "solution" the next time, then don't post it up and waste other ppl's time. Incorrect or erroneous "solutions" can also completely mislead or misguide ppl, who sincerely want to learn, into misinformation and a wrong grasp of concepts.
And that is a risk you should not take upon the shoulders of others.

 

[SeDaTeD]

so until or unless you are very confident of your "solution" the next time, then don't post it up and waste other ppl's time. Incorrect or erroneous "solutions" can also completely mislead or misguide ppl, who sincerely want to learn, into misinformation and a wrong grasp of concepts.
And that is a risk you should not take upon the shoulders of others.

I think you're being a bit too harsh there. If that was the case, nobody would post a solution. This is a thread for discussion or ideas. Take note: people are allowed to be wrong! You shouldn't come down on them so hard if they can't come up a non-erroneous solution. I don't think it wastes other people's times, in fact it's good that people are able to discuss maths. It's part of learning.

If you keep attacking every suggestion that comes up then nobody would want to try their hand at the question, or extra-curricular maths in general. It's good that you point out errors in arguement, but to suggest that one should not post their solutions as it wastes people's time and misleads them is going way over the top.

 

[Templar]

No, I believe who_loves_maths' response is appropriate. casebash is not reading everything and clearly believes he's too good for the rest of us. The only reason I haven't made any comments on the validity of his methods is that I am somewhat tired of trying to correct his errors only for him to say then it's still easy and come up with another invalid one.

Even permitting the use of coordinate geometry, you have a ruler, not a calculator, and I doubt anyone can mentally solve trig for angles with transcendental solutions.

 

[who_loves_maths]

Originally Posted by casebash
Ah, got it, a real geometric proof. Find the midpoints of the triangle. Draw a line through two of them to get a line parrallel to the base. Join both the bisection of this and the bisection of the base to get a line through the center. Bisect that.

Ahhh.... now that's what i'm talking about! Yes, you've got it, that's the right solution. Finally someone comes up with it - i was beginning to think (esp. after your penultimate post) that no-one here will come to this solution.
but yes, well done casebash. very good work.

i had two solutions to this problem when i started the thread, hoping to see if someone can come up with a better or more elegant solution. your method here is one of the two i have ... but right now, it looks unlikely that there's going to be a third one found anytime soon.
a secondary reason for starting this thread for me was to acknowledge the simplicity of the solutions to a somewhat seemingly "difficult" problem here. who would've imagined a problem similar to the Jacob Steiner one {which is impossible to solve}, except with the straightedge being given calibrations, would require only a very elementary proof reliant on the simple fact that a line connecting the midpoints of two sides of a triangle is parallel to its base.

but anyhow, obviously you were good enough to spot it ... so once again, well done!

so now to SeDaTeD: Nope, a solution is not impossible, which you said you suspected in your last post.


P.S. the second solution to this problem, for those still interested, lies on another simple geometric fact of the circle - that the chord joining the midpoints of any other two chords in the circle is ALWAYS longer than either of these chords.
ie. by continually contructing longer chords using this fact in a structured way will produce two successive 'chords' of equal length - when this happens, these two equal chords are diameters of the circle, and since there are two of them then the point of their intersection is indeed the centre of the circle.

 

[Templar]

by continually contructing longer chords using this fact in a structured way will produce two successive 'chords' of equal length - when this happens, these two equal chords are diameters of the circle, and since there are two of them then the point of their intersection is indeed the centre of the circle.

Have you considered that it might take infinite number of steps, and that even if they do become diameters you wouldn't know?

 

[SeDaTeD]

woops, posted after that last post. this one's irrelevent now.

 

[who_loves_maths]

Originally Posted by Templar
Have you considered that it might take infinite number of steps.

uhhuh, that's why it's the second solution, not my first.
i first found the same solution as casebash you see, and then i went and tried to look for more out there {as i said in my last post, that was the reason for starting this thread}. and the result was this second one.
so in the state of being semi-inspired, i started this thread, hoping that someone out there might found more solutions to this particular problem.

and my uncertainty about how long it might take to undertake this second solution is why i made that final post congradulating casebash on finding the solution - since he found the first one, which is 100% correct.

if he'd thought of the second one, then i might have had some reservations in posting what i did in my last post.

but anyhow, like you said, the uncertainty over this area of the second method is still a problem... but i am currently in the process of trying to decide/show whether or not how long it'll take is completely random and indeterminate or whether all starting points will eventually end up in the diameter...
{which, btw, out of interest, is another problem for you to solve if you like to }

so i might get back to you if i (i hope) eventually find out about this.

 

[who_loves_maths]

Originally Posted by Templar
... and that even if they do become diameters you wouldn't know?

sorry Templar, but here you are wrong...
read what i typed in my post to casebash again (bolded below):

Originally Posted by who_loves_maths
P.S. the second solution to this problem, for those still interested, lies on another simple geometric fact of the circle - that the chord joining the midpoints of any other two chords in the circle is ALWAYS longer than either of these chords.
ie. by continually contructing longer chords using this fact in a structured way will produce two successive 'chords' of equal length - when this happens, these two equal chords are diameters of the circle, and since there are two of them then the point of their intersection is indeed the centre of the circle.

 

[Templar]

It might still be an infinitely converging method.